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(B) The relationship between the standard cell potential $(E^o_{cell})$ and the equilibrium constant $(K_c)$ is given by the Nernst equation at $298 \

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$+0.2955$

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(B) The relationship between the standard cell potential $(E^o_{cell})$ and the equilibrium constant $(K_c)$ is given by the Nernst equation at $298 \ K$:
$E^o_{cell} = \frac{0.0591}{n} \log K_c$
First,determine the number of electrons $(n)$ transferred in the reaction:
$2X_2 	o 2X_2^{3+} + 6e^-$
$3Y^{2+} + 6e^- 	o 3Y$
Thus,$n = 6$.
Given $K_c = 10^{30}$ and $n = 6$:
$E^o_{cell} = \frac{0.0591}{6} \log(10^{30})$
$E^o_{cell} = \frac{0.0591}{6} 	imes 30$
$E^o_{cell} = 0.0591 	imes 5$
$E^o_{cell} = 0.2955 \ V$

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^o$ for the cell at $298 \ K$.
$2 X_2(s)+3 Y^{2+}(a q) \rightarrow 2 X_2^{3+}(a q)+3 Y(s)$

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The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^o$ for the cell at $298 \ K$. $2 X_2(s)+3 Y^{2+}(a q) \rightarrow 2 X_2^{3+}(a q)+3 Y(s)$