[Given:
$\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \quad \text{E}^\circ_{\text{red}} = +0.34 \text{ V}$
$\text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \quad \text{E}^\circ_{\text{red}} = +1.23 \text{ V}$
Molar mass of Cu = $63.54 \text{ g mol}^{-1}$
Molar mass of $\text{O}_2$ = $32 \text{ g mol}^{-1}$
Faraday Constant = $96500 \text{ C mol}^{-1}$
Molar volume at STP = 22.4 L]
Electrolysis: Calculating Total Oxygen Evolved
This problem involves calculating the total volume of oxygen ($O_2$) evolved during the electrolysis of an acidic copper(II) solution ($\text{Cu}^{2+}$). The process occurs in two conceptual phases:
- Phase 1: Deposition of Copper metal ($\text{Cu}$).
- Phase 2: Evolution of Oxygen gas ($\text{O}_2$).
We assume the given current of 600 mA ($0.6$ A) was applied throughout the process, including during copper deposition and subsequent oxygen evolution.
Phase 1: Copper Deposition
First, calculate the amount of charge passed to deposit 300 mg of Copper.
- Calculate moles of Cu:
Mass of Cu = 300 mg = $0.300$ g
Molar mass of Cu = $63.54$ g mol$^{-1}$
Moles of Cu = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{0.300 \, \text{g}}{63.54 \, \text{g mol}^{-1}} \approx 0.004721$ mol
- Calculate moles of electrons for Cu deposition:
The reduction reaction is $\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}$.
Moles of electrons ($\text{e}^-$) = $2 \times$ Moles of Cu = $2 \times 0.004721 \approx 0.009443$ mol
- Calculate charge ($Q$) for Cu deposition:
Faraday constant (F) = $96500$ C mol$^{-1}$
$Q_{\text{Cu}} = \text{Moles of e}^- \times F = 0.009443 \, \text{mol} \times 96500 \, \text{C mol}^{-1} \approx 911.25$ C
- Calculate time ($t$) for Cu deposition:
Current (I) = $0.6$ A
$t_{\text{Cu}} = \frac{Q_{\text{Cu}}}{I} = \frac{911.25 \, \text{C}}{0.6 \, \text{A}} \approx 1518.75$ seconds
- Calculate O$_2$ evolved during Cu deposition:
The oxidation of water produces $\text{O}_2$: $2\text{H}_2\text{O} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^-$.
Moles of $\text{O}_2$ = $\frac{\text{Moles of e}^-}{4} = \frac{0.009443 \, \text{mol}}{4} \approx 0.002361$ mol
- Calculate volume of O$_2$ at STP during Phase 1:
Molar volume at STP = $22.4$ L mol$^{-1}$
Volume $V_{O_2, \text{Cu}} = \text{Moles of } O_2 \times \text{Molar volume at STP}$
$V_{O_2, \text{Cu}} = 0.002361 \, \text{mol} \times 22.4 \, \text{L mol}^{-1} \approx 0.05291$ L
$V_{O_2, \text{Cu}} = 0.05291 \times 1000$ mL $\approx 52.91$ mL
Phase 2: Oxygen Evolution
Calculate the charge passed and oxygen evolved during the additional 28 minutes.
- Calculate time in seconds:
Additional time = 28 minutes = $28 \times 60 = 1680$ seconds
- Calculate charge ($Q$) passed in Phase 2:
$Q_{\text{add}} = I \times t = 0.6 \, \text{A} \times 1680 \, \text{s} = 1008$ C
- Calculate moles of electrons passed in Phase 2:
Moles of $\text{e}^-$ ($\text{e}^-_{\text{add}}$) = $\frac{Q_{\text{add}}}{F} = \frac{1008 \, \text{C}}{96500 \, \text{C mol}^{-1}} \approx 0.010446$ mol
- Calculate moles of O$_2$ evolved in Phase 2:
Moles of $\text{O}_2$ = $\frac{\text{Moles of e}^-_{\text{add}}}{4} = \frac{0.010446 \, \text{mol}}{4} \approx 0.002611$ mol
- Calculate volume of O$_2$ at STP during Phase 2:
$V_{O_2, \text{add}} = \text{Moles of } O_2 \times \text{Molar volume at STP}$
$V_{O_2, \text{add}} = 0.002611 \, \text{mol} \times 22.4 \, \text{L mol}^{-1} \approx 0.05853$ L
$V_{O_2, \text{add}} = 0.05853 \times 1000$ mL $\approx 58.53$ mL
Total Oxygen Volume
Sum the volumes evolved in both phases.
- Calculate total volume of O$_2$:
Total Volume $V_{O_2, \text{total}} = V_{O_2, \text{Cu}} + V_{O_2, \text{add}}$
$V_{O_2, \text{total}} \approx 52.91 \text{ mL} + 58.53 \text{ mL} = 111.44$ mL
- Round to the nearest integer:
The total volume of oxygen evolved is approximately 111 mL.
Similar Questions
- X and Y are the number of electrons involved, respectively during the oxidation of $\text{I}^-$ to $\text{I}_2$ and $\text{S}^{2-}$ to S by acidified $\text{K}_2\text{Cr}_2\text{O}_7$. The value of X + Y is ______.
- Consider two Group IV metal ions $\text{X}^{2+}$ and $\text{Y}^{2+}$.
A solution containing 0.01 M $\text{X}^{2+}$ and 0.01 M $\text{Y}^{2+}$ is saturated with $\text{H}_2\text{S}$. The pH at which the metal sulphide YS will form as a precipitate is ______. (Nearest integer)
(Given: $\text{K}_{sp}(\text{XS}) = 1 \times 10^{-22}$ at $25^\circ\text{C}$, $\text{K}_{sp}(\text{YS}) = 4 \times 10^{-16}$ at $25^\circ\text{C}$, $[\text{H}_2\text{S}] = 0.1\text{M}$ in solution, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$) - The half-life of ${}^{65}\text{Zn}$ is 245 days. After x days, 75% of original activity remained. The value of $x$ in days is _______. (Nearest integer)
(Given: $\log 3 = 0.4771$ and $\log 2 = 0.3010$) - Molar conductivity of a weak acid HQ of concentration $0.18 \text{ M}$ was found to be $1/30$ of the molar conductivity of another weak acid HZ with concentration of $0.02 \text{ M}$. If $\lambda^\circ_{Q^-}$ happened to be equal with $\lambda^\circ_{Z^-}$, then the difference of the $\text{pK}_a$ values of the two weak acids ($\text{pK}_a\text{(HQ)} - \text{pK}_a\text{(HZ)}$) is _______ (Nearest integer).
[Given: degree of dissociation ($\alpha$) < < 1 for both weak acids, $\lambda^\circ$ : limiting molar conductivity of ions] - The first and second ionization constants of $\text{H}_2\text{X}$ are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. The concentration of $\text{X}^{2-}$ in $0.1\text{ M}$ $\text{H}_2\text{X}$ solution is __________ $\times 10^{-15}\text{ M}$. (Nearest Integer)
- The osmotic pressure of a living cell is $12\text{ atm}$ at $300\text{ K}$. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is __________ $\text{g L}^{-1}$. (Nearest integer)
Given : $\text{R} = 0.08\text{ L atm K}^{-1}\text{ mol}^{-1}$
Assume complete dissociation of $\text{NaCl}$
(Given : Molar mass of $\text{Na}$ and $\text{Cl}$ are $23$ and $35.5\text{ g mol}^{-1}$ respectively.) - $MX$ is a sparingly soluble salt that follows the given solubility equilibrium at $298\text{ K}$.
$\text{MX (s)} \rightleftharpoons \text{M}^+\text{ (aq)} + \text{X}^-\text{ (aq)}; \quad \text{K}_{\text{sp}} = 10^{-10}$
If the standard reduction potential for $\text{M}^+\text{(aq)} \xrightarrow{+\text{e}^-} \text{M(s)}$ is $\left(\text{E}^\ominus_{\text{M}^+/\text{M}}\right) = 0.79\text{ V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $\text{E}^\ominus_{\text{X}^-/\text{MX (s)}/\text{M}}$ is __________ $\text{mV}$. (nearest integer)
[Given : $\frac{2.303\text{ RT}}{\text{F}} = 0.059\text{ V}$] - For strong electrolyte $\Lambda_{\text{m}}$ increases slowly with dilution and can be represented by the equation $\Lambda_{\text{m}} = \Lambda_{\text{m}}^{\circ} - \text{A}\text{c}^{1/2}$
Molar conductivity values of the solutions of strong electrolyte AB at $18^{\circ}\text{C}$ are given below :$\text{c } [\text{mol L}^{-1}]$ $0.04$ $0.09$ $0.16$ $0.25$ $\Lambda_{\text{m}} \ [\text{S cm}^{2}\text{ mol}^{-1}]$ $96.1$ $95.7$ $95.3$ $94.9$
The value of constant A based on the above data $[\text{in S cm}^{2}\text{ mol}^{-1} / (\text{mol}/\text{L})^{1/2}]$ unit is _________. - $\text{A} \longrightarrow \text{B} \text{ (first reaction)}$
$\text{C} \longrightarrow \text{D} \text{ (second reaction)}$
Consider the above two first-order reactions. The rate constant for first reaction at $500\text{ K}$ is double of the same at $300\text{ K}$. At $500\text{ K}$, $50\%$ of the reaction becomes complete in $2\text{ hour}$. The activation energy of the second reaction is half of that of first reaction. If the rate constant at $500\text{ K}$ of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at $300\text{ K}$ is _________ $\times 10^{-1}\text{hour}^{-1}$ (nearest integer). For the thermal decomposition of reactant $\text{AB(g)}$, the following plot is constructed.
The half life of the reaction is '$x$' min.
$x = $ ___________ min. (Nearest integer)
Important Questions from Physical Chemistry
- X and Y are the number of electrons involved, respectively during the oxidation of $\text{I}^-$ to $\text{I}_2$ and $\text{S}^{2-}$ to S by acidified $\text{K}_2\text{Cr}_2\text{O}_7$. The value of X + Y is ______.
- Consider two Group IV metal ions $\text{X}^{2+}$ and $\text{Y}^{2+}$.
A solution containing 0.01 M $\text{X}^{2+}$ and 0.01 M $\text{Y}^{2+}$ is saturated with $\text{H}_2\text{S}$. The pH at which the metal sulphide YS will form as a precipitate is ______. (Nearest integer)
(Given: $\text{K}_{sp}(\text{XS}) = 1 \times 10^{-22}$ at $25^\circ\text{C}$, $\text{K}_{sp}(\text{YS}) = 4 \times 10^{-16}$ at $25^\circ\text{C}$, $[\text{H}_2\text{S}] = 0.1\text{M}$ in solution, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$) - The half-life of ${}^{65}\text{Zn}$ is 245 days. After x days, 75% of original activity remained. The value of $x$ in days is _______. (Nearest integer)
(Given: $\log 3 = 0.4771$ and $\log 2 = 0.3010$) - Molar conductivity of a weak acid HQ of concentration $0.18 \text{ M}$ was found to be $1/30$ of the molar conductivity of another weak acid HZ with concentration of $0.02 \text{ M}$. If $\lambda^\circ_{Q^-}$ happened to be equal with $\lambda^\circ_{Z^-}$, then the difference of the $\text{pK}_a$ values of the two weak acids ($\text{pK}_a\text{(HQ)} - \text{pK}_a\text{(HZ)}$) is _______ (Nearest integer).
[Given: degree of dissociation ($\alpha$) < < 1 for both weak acids, $\lambda^\circ$ : limiting molar conductivity of ions] - The first and second ionization constants of $\text{H}_2\text{X}$ are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. The concentration of $\text{X}^{2-}$ in $0.1\text{ M}$ $\text{H}_2\text{X}$ solution is __________ $\times 10^{-15}\text{ M}$. (Nearest Integer)
