How would you determine the standard electrode potential of the $M g^{2 +}$ $| M g$ ?

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Solution

To determine the electrode potential of $m g^{+ 2} / m g^{1}$ we use hydrogen electrode (anode)
The standard hydrogen electrode potential is zero.
$∴ E^{o} c e l l = E^{o} r i g h t + E^{o} l e f t$
$E^{o} c e l l = E^{o} m g^{+ 2} / m g^{- 1} E^{o} n_{2} / n_{1}$
$E^{o} m g^{+ 2} / m g = E^{o} c e l l$

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Similar questions

E_{M} g^{2} + | M g

E_{M} g^{2} + | M g^{Θ}

\frac{0.059}{2} l o g \frac{1}{[M g^{2 +}]}

E_{M g^{2 +} | M g}

l o g [M g^{2 +}]

E_{\frac{M g^{2 +}}{M g}} = E_{\frac{M g^{2 +}}{M g}}^{\circ} - \frac{0.059}{2} l o g (\frac{1}{M g^{2 +}})

E_{\frac{M g^{2 +}}{M g}} v s l o g [M g^{2 +}]

Given the standard electrode potentials,

$\frac{K^{+}}{K} = - 2.93 V,$

$\frac{A g^{+}}{A g} = 0.80 V$

$\frac{H g^{2 +}}{H g} = 0.79 V,$

$\frac{M g^{2 +}}{M g} = - 2.37 V$

$\frac{C r^{3 +}}{C r} = - 0.74 V$

Arrange these metals in their increasing order of reducing power.

K^{+} / K = - 2.93 V, A g^{+} / A g = 0.80 V,

H g^{2 +} / H g = 0.79 V, M g^{2 +} / M g = - 2.37 V, C r^{3 +} / C r = - 0.74 V

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