The e.m.f. of the cell in which the following | vedclass

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Question

(B) The cell reaction is: $Zn_{(s)} + Ni^{2+}(a = 1.0) \rightarrow Zn^{2+}(a = 10) + Ni_{(s)}$.Here,$Zn$ undergoes oxidation $(Zn \rightarrow Zn^{2+}

Vedclass · Accepted Answer

$0.54$

Vedclass · Answer

(B) The cell reaction is: $Zn_{(s)} + Ni^{2+}(a = 1.0) ightarrow Zn^{2+}(a = 10) + Ni_{(s)}$.Here,$Zn$ undergoes oxidation $(Zn ightarrow Zn^{2+} + 2e^-)$,acting as the anode,and $Ni^{2+}$ undergoes reduction $(Ni^{2+} + 2e^- ightarrow Ni)$,acting as the cathode.Using the Nernst equation at $298 \ K$:$E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Ni^{2+}]}$Given:$E_{cell} = 0.5105 \ V$$n = 2$$[Zn^{2+}] = 10$$[Ni^{2+}] = 1.0$Substituting the values:$0.5105 = E_{cell}^{o} - \frac{0.0591}{2} \log \frac{10}{1}$$0.5105 = E_{cell}^{o} - 0.02955 	imes 1$$E_{cell}^{o} = 0.5105 + 0.02955 = 0.54005 \ V \approx 0.54 \ V$.Thus,the standard e.m.f. of the cell is $0.54 \ V$.

The e.m.f. of the cell in which the following reaction $Zn_{(s)} + Ni^{2+}(a = 1.0) \rightleftharpoons Zn^{2+}(a = 10) + Ni_{(s)}$ occurs,is found to be $0.5105 \ V$ at $298 \ K$. The standard e.m.f. of the cell is ............ $V$.

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