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(B) The cell reaction is: $Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$Using the Nernst equation: $E_{cell} = E_{cell}^{\

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$10.4115$

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(B) The cell reaction is: $Ni_{(s)} + 2Ag^{+}(0.001 \ M) ightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$Using the Nernst equation: $E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^2}$Here,$n = 2$,$[Ni^{2+}] = 10^{-3} \ M$,and $[Ag^{+}] = 10^{-3} \ M$.$E_{cell} = 10.5 - \frac{0.059}{2} \log \frac{10^{-3}}{(10^{-3})^2}$$E_{cell} = 10.5 - \frac{0.059}{2} \log \frac{10^{-3}}{10^{-6}}$$E_{cell} = 10.5 - \frac{0.059}{2} \log 10^3$$E_{cell} = 10.5 - \frac{0.059}{2} 	imes 3$$E_{cell} = 10.5 - 0.0885 = 10.4115 \ V$

Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):
$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$
(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)

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Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)