electrochemistry

Question: .$Zn+C{{u}^{2+}}(aq) \rightleftharpoons Cu+Z{{n}^{2+}}(aq)$, reaction quotient is $Q=\frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}$. Variation of ${{E}_{\text{cell}}}$ with log Q is of the type with OA = 1.10 V. ${{E}_{\text{cell}}}$ will be 1.1591 V when



1) $\frac{[C{{u}^{2+}}]}{[Z{{n}^{2+}}]}=0.01$
2) $\frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}=0.01$
3) $\frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}=0.1$
4) $\frac{[Z{{n}^{2+}}]}{[C{{u}^{2+}}]}=1$
Solution: Explanation: No Explanation
Electrode potential Ecell Nernst equation and ECS

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