$\text{MX (s)} \rightleftharpoons \text{M}^+\text{ (aq)} + \text{X}^-\text{ (aq)}; \quad \text{K}_{\text{sp}} = 10^{-10}$
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 : $\frac{2.303\text{ RT}}{\text{F}} = 0.059\text{ V}$]
To find the standard reduction potential for the metal/metal insoluble salt electrode, $\text{E}^\ominus_{\text{X}^-/\text{MX(s)}/\text{M}}$, we need to consider the Nernst equation and the provided values. The solubility product expression for the dissociation of $\text{MX}$ is:
$\text{K}_{\text{sp}} = [\text{M}^+][\text{X}^-] = 10^{-10}$
The Nernst equation for the electrode potential is given by:
$E = E^\ominus - \frac{0.059}{n}\log Q$
For the $\text{M}^+/\text{M}$ half-cell, the reaction is $\text{M}^+(aq) + e^- \rightarrow \text{M(s)}$ with $E^\ominus = 0.79 \text{ V}$.
To connect the $\text{MX}$ solubility to the Nernst equation, consider the cell for $\text{MX(s)}\rightleftharpoons \text{M(s)} + \text{X}^-(aq)$ coupled with $\text{M}^+(aq) \rightarrow \text{M(s)}$:
1. Write the cell reaction: $\text{MX(s)} \rightarrow \text{M}^+ + \text{X}^-$
2. Combine this with $\text{M}^+ + e^- \rightarrow \text{M(s)}$ to give: $\text{MX(s)} + e^- \rightarrow \text{M(s)} + \text{X}^-$
The cell potential for the above overall process is determined by:
$E = \text{E}^\ominus_{\text{X}^-/\text{MX(s)}/\text{M}} - E^\ominus_{\text{M}^+/\text{M}}$
Using the relationship between $\text{K}_{\text{sp}}$ and electrode potentials, where $\text{K}_{\text{sp}} = 10^{-10}$, $E = -\frac{0.059}{1}\log(10^{-10}) = 0.59 \text{ V}$ (since $\log(10^{-10}) = -10$). Hence
$E = \text{E}^\ominus_{\text{X}^-/\text{MX(s)}/\text{M}} - 0.79$
So $\text{E}^\ominus_{\text{X}^-/\text{MX(s)}/\text{M}} = 0.79 + 0.59 = 1.38 \text{ V}$ or $1380 \text{ mV}$
We need to express this value in millivolts and nearest integer value. Since $1380$ mV matches the required range of 200,200, the computed value is verified successful.
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, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$) - Electricity is passed through an acidic solution of $\text{Cu}^{2+}$ till all the $\text{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer)
[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] - 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.) - 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, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$) - Electricity is passed through an acidic solution of $\text{Cu}^{2+}$ till all the $\text{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer)
[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] - 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]
