For the cell reaction Zn(s) + Cu^{2+}(aq) (1. | vedclass

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(C) Using the Nernst equation at $298 \, K$:$E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Given $n = 2$,$E_{cell} = 1.3 \

Vedclass · Accepted Answer

$1.27$

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(C) Using the Nernst equation at $298 \, K$:$E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Given $n = 2$,$E_{cell} = 1.3 \, V$,$[Zn^{2+}] = 0.1 \, M$,and $[Cu^{2+}] = 1.0 \, M$.Substituting the values:$1.3 = E^o_{cell} - \frac{0.0591}{2} \log \frac{0.1}{1.0}$$1.3 = E^o_{cell} - 0.02955 \log(10^{-1})$$1.3 = E^o_{cell} + 0.02955 \log(10)$Since $\log(10) = 1$:$1.3 = E^o_{cell} + 0.02955$$E^o_{cell} = 1.3 - 0.02955 = 1.27045 \, V \approx 1.27 \, V$.

For the cell reaction $Zn(s) + Cu^{2+}(aq) (1.0 \, M) \to Cu(s) + Zn^{2+}(aq) (0.1 \, M)$,the measured $e.m.f.$ at $25 \, ^oC$ is $1.3 \, V$. Calculate the $E^o$ value for the cell reaction. (in $, V$)

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