What is E_{cell} (in V) of the following cell | vedclass

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(D) The cell reaction is: $Zn_{(s)} + Ni^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ni_{(s)}$The standard cell potential is: $E^{\ominus}_{cell} = E^{\o

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$0.54$

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(D) The cell reaction is: $Zn_{(s)} + Ni^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ni_{(s)}$The standard cell potential is: $E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode} = E^{\ominus}_{Ni^{2+}/Ni} - E^{\ominus}_{Zn^{2+}/Zn}$$E^{\ominus}_{cell} = -0.25 \ V - (-0.76 \ V) = 0.51 \ V$Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} - \frac{0.06}{n} \log \frac{[Zn^{2+}]}{[Ni^{2+}]}$Here,$n = 2$,$[Zn^{2+}] = 0.01 \ M$,and $[Ni^{2+}] = 0.1 \ M$$E_{cell} = 0.51 - \frac{0.06}{2} \log \frac{0.01}{0.1}$$E_{cell} = 0.51 - 0.03 \log(0.1)$$E_{cell} = 0.51 - 0.03(-1) = 0.51 + 0.03 = 0.54 \ V$

What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

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What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$