Which graph correctly correlates E_{cell} as | vedclass

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(B) The Nernst equation for the given cell reaction is:$E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Here,$n = 2$,so the e

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(B) The Nernst equation for the given cell reaction is:$E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Here,$n = 2$,so the equation becomes:$E_{cell} = 1.10 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$This is in the form of a linear equation $y = mx + c$,where $y = E_{cell}$,$x = log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$c = 1.10$,and the slope $m = -\frac{0.059}{2} = -0.0295$.Since the slope is negative,the graph of $E_{cell}$ versus $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$ will be a straight line with a negative slope and a $y$-intercept of $1.10 \, V$.

Which graph correctly correlates $E_{cell}$ as a function of concentrations for the cell (for different values of $M$ and $M'$):-
$Zn_{(s)} + Cu^{2+}(M) \to Zn^{2+}(M') + Cu_{(s)};$ $E^o_{cell} = 1.10 \, V$
$X$-axis : $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$Y$-axis : $E_{cell}$

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Which graph correctly correlates $E_{cell}$ as a function of concentrations for the cell (for different values of $M$ and $M'$):-$Zn_{(s)} + Cu^{2+}(M) \to Zn^{2+}(M') + Cu_{(s)};$ $E^o_{cell} = 1.10 \, V$$X$-axis : $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$Y$-axis : $E_{cell}$