If E^{circ}(Mg^{+2}_{(aq)} mid Mg_{(s)}) = -2 | vedclass

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Question

(C) The given reaction is the oxidation of $Mg$ to $Mg^{+2}$. The standard reduction potential is $E^{\circ}(Mg^{+2} \mid Mg) = -2.37 \ V$. Therefore,

Vedclass · Accepted Answer

$+2.4292 \ V$

Vedclass · Answer

(C) The given reaction is the oxidation of $Mg$ to $Mg^{+2}$. The standard reduction potential is $E^{\circ}(Mg^{+2} \mid Mg) = -2.37 \ V$. Therefore,the standard oxidation potential is $E^{\circ}_{ox} = -(-2.37 \ V) = +2.37 \ V$. Using the Nernst equation for the oxidation half-cell: $E_{ox} = E^{\circ}_{ox} - \frac{0.0591}{n} \log [Mg^{+2}]$. Here,$n = 2$ and $[Mg^{+2}] = 0.01 \ M = 10^{-2} \ M$. Substituting the values: $E_{ox} = 2.37 - \frac{0.0591}{2} \log(10^{-2})$. $E_{ox} = 2.37 - 0.02955 	imes (-2)$. $E_{ox} = 2.37 + 0.0591 = 2.4291 \ V$. Rounding to four decimal places,we get $2.4292 \ V$.

If $E^{\circ}(Mg^{+2}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$. What is the potential for $Mg_{(s)} \rightarrow Mg^{+2}_{(0.01 \ M)} + 2 \overline{e}$ at $298 \ K$?

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