For the cell Zn_{(s)} | Zn^{2+} (0.1 M) || F | vedclass

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Question

(A) The cell reaction is: $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$Using the Nernst equation: $E_{cell} = E^{\circ}_{ce

Vedclass · Accepted Answer

$10^{0.32/0.0295}$

Vedclass · Answer

(A) The cell reaction is: $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Fe^{2+}]}$Given $E_{cell} = 0.2905 \ V$,$n = 2$,$[Zn^{2+}] = 0.1 \ M$,$[Fe^{2+}] = 0.01 \ M$:$0.2905 = E^{\circ}_{cell} - \frac{0.059}{2} \log \frac{0.1}{0.01}$$0.2905 = E^{\circ}_{cell} - 0.0295 \log(10)$$0.2905 = E^{\circ}_{cell} - 0.0295 \implies E^{\circ}_{cell} = 0.32 \ V$At equilibrium,$E_{cell} = 0$,so $E^{\circ}_{cell} = \frac{0.059}{n} \log K_c$$0.32 = \frac{0.059}{2} \log K_c$$0.32 = 0.0295 \log K_c$$\log K_c = \frac{0.32}{0.0295}$$K_c = 10^{0.32/0.0295}$

For the cell $Zn_{(s)} | Zn^{2+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe_{(s)}$ at $298 \ K$,$E_{cell} = 0.2905 \ V$. What is the equilibrium constant $(K_c)$ for the reaction $Zn_{(s)} + Fe^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Fe_{(s)}$?

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