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Electrochemical Cells, Galvanic Cells

The emf of a ...

Question

The emf of a Daniel cell at 298 K is $E_{1}$ concentration of $Z n S O_{4}$ is 0.1 M and that $C u S O_{4}$ is 0.01 M, the emf changed to $E_{2}$ .What is the relationship between $E_{1}$ and $E_{2}$

A

E_{2} = 0 \neq E_{1}

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B

E_{1} > E_{2}

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C

E_{2} > E_{1}

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D

E_{1} = E_{2}

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Solution

The correct option is C $E_{2} > E_{1}$
$\begin{matrix} Given ϵ_{0} = ϵ_{1} we know that at temperature 298 k . ϵ = ϵ_{0} - \frac{0.06}{n} log \frac{[2 n^{+ 2}]}{[c u^{+ 2}]} \end{matrix}$
$\begin{matrix} ϵ_{2} = ϵ_{1} - \frac{0.06}{2} log \frac{0.01}{1} ϵ_{2} = ϵ_{1} - \frac{0.06}{2} (- 2) ϵ_{2} = ϵ_{1} + 0.06 ϵ_{2} - ϵ_{1} = 0.06 ϵ_{2} - ϵ_{1} > 0 ϵ_{2} > ϵ_{1} option c is correct \end{matrix}$

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