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Given $E_{{Zn}^{2} / Zn}^{\circ} = - 0 .76 V; E_{{Ni}^{2} + / Ni}^{0} = - 0 .25 V$ .
Calculate the EMF of the cell, where the following reaction is taking place.
${Zn}_{(s)} + {Ni}_{(aq)}^{2 +} ⟶ {Zn}_{(aq)}^{2 +} + {Ni}_{(s)}$

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$- 0.51 V$

$1.01 V$

$0.25 V$

$0.51 V$

answer is A.

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Detailed Solution

A cell reaction is possible, if $E_{cell}$ is positive.
Since $- 0.25 V > - 0.76 V,$
So: $\begin{array}{l} E_{cell} = E_{red}^{\circ} of Ni - E_{red}^{\circ} of Zn \\ = - 0 .25 V - (- 0 .76 V) = + 0 .51 V \end{array}$

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