The equilibrium constant for the reaction: Zn(s) + Sn^2+(aq) Zn^2+(aq) + Sn(s) is 1 10^20 at 298 K. The magnitude of standard electrode potential of Sn^2+/Sn if E^ Zn^2+/Zn = -0.76 V is 10^-2 V…

Given: K = 1 10^20 at 298 K and E^ Zn^2+/Zn = -0.76 V. Find: The magnitude of E^ Sn^2+/Sn in the form 10^-2 V. The equilibrium constant is related to the electrode potentials: G^ = -nF E^ = -2.303RT K Hence, E^ cell = 0.0592 K Substituting K = 1 10^20: E^ cell = 0.0592 (1 10^20) = 0.059 10 = 0.59 V Using E^ cell = E^ Sn^2+/Sn - E^ Zn^2+/Zn we get 0.59 = E^ Sn^2+/Sn - (-0.76) E^ Sn^2+/Sn = 0.59 - 0.76 = 0.17 V = 17 10^-2 V Therefore, the required nearest integer is 17.

The equilibrium constant for the reaction: Zn(s) + Sn^2+(aq) Zn^2… — Solution | JEE Chemistry

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The equilibrium constant for the reaction: $Zn(s) + Sn^{2+}(aq) \rightleftharpoons Zn^{2+}(aq) + Sn(s)$

is $1 \times 10^{20}$ at $298 \, \text{K}$ . The magnitude of standard electrode potential of $Sn^{2+}/Sn$ if $E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$ is $\ldots \times 10^{-2} \, V$ (Nearest integer).

Answer

Correct answer:17

Step-by-step solution

Standard Method

Given: $K = 1 \times 10^{20}$ at $298 \, \text{K}$ and $E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$ .

Find: The magnitude of $E^\circ_{Sn^{2+}/Sn}$ in the form $\ldots \times 10^{-2} \, V$ .

The equilibrium constant is related to the electrode potentials:

\Delta G^\circ = -nF E^\circ = -2.303RT \log K

Hence,

E^\circ_{\text{cell}} = \frac{0.059}{2} \log K

Substituting $K = 1 \times 10^{20}$ :

E^\circ_{\text{cell}} = \frac{0.059}{2} \log (1 \times 10^{20}) = 0.059 \times 10 = 0.59 \, \text{V}

Using

E^\circ_{\text{cell}} = E^\circ_{Sn^{2+}/Sn} - E^\circ_{Zn^{2+}/Zn}

we get

0.59 = E^\circ_{Sn^{2+}/Sn} - (-0.76)

E^\circ_{Sn^{2+}/Sn} = 0.59 - 0.76 = 0.17 \, \text{V} = 17 \times 10^{-2} \, \text{V}

Therefore, the required nearest integer is 17.

Common mistakes

Using $E^\circ_{\text{cell}} = E^\circ_{Zn^{2+}/Zn} - E^\circ_{Sn^{2+}/Sn}$ in the wrong order. This reverses the sign of the cell potential. Use cathode minus anode exactly as written for the reaction.
Forgetting that $n = 2$ for the reaction. Since two electrons are transferred, the factor in $E^\circ_{\text{cell}} = \frac{0.059}{n} \log K$ must be $\frac{0.059}{2}$ , not $0.059$ .
Calculating $\log(1 \times 10^{20})$ incorrectly as $20$ . Because $\log 1 = 0$ , the value is only $20$ , and after multiplying by $\frac{0.059}{2}$ the result is $0.59 \, \text{V}$ .

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