# Question 65f39

Mar 2, 2017

$1 A g + \frac{1}{2} C {a}^{+ 2} \rightarrow \frac{1}{2} C a + 1 A {g}^{+}$

$2 A g \rightarrow A {g}^{+} + 2 {e}^{-}$
We have given the ${E}_{\text{red}} = 0.766 V$
Therefore Eox = -Ered
Eox = -0.766V

Ca^(+2) + 2e^-) rarr Ca 
${E}_{\text{red}} = - 2.76 V$

${E}^{o} \text{cell" = E_"red" + E_"ox}$

$- 2.76 V + - 0.766 V = - 3.526 V$

If the ${E}^{o} c e l l$ is negative the reaction is non-spontaneous

As the concentration $C {a}^{2 +} \mathmr{and} A {g}^{+}$ is 0.1M and not 1M you will have to solve Nerst Equation .

$E c e l l = {E}^{o} - \text{RT"/n} \log Q$

-3.526V - 0.0591/2 log Q

$Q = {0.1}^{0.5} / {0.1}^{1} = \frac{0.1}{0.1} ^ 2$

$E c e l l = - 3.526 V - \frac{0.0591}{2} \log \left(\frac{0.1}{0.1} ^ 2\right)$

-3.526V - 0.02955 = -3.55555V

You can also solve by following the ln method but this the easy way

Now as the ${E}_{\text{cell}}$ is negative the reaction will not occur as written as ${E}_{\text{cell}}$ is negative for non-spontaneous reaction and positive for spontaneous reaction .

The spontaneous reaction opposite of this non spontaneous reaction would be oxidation of Ca and reduction of Ag+

$C a \rightarrow C {a}^{+ 2} + 2 {e}^{-}$
${E}_{\text{ox}} = 2.76 V$

2Ag^+ + 2e^-) rarr Ag#
${E}_{\text{red}} = 0.766 V$

${E}^{o} c e l l = 3.526 V$

Therefore this reaction is a spontaneous one.

The Ecell would remain the same since the concentration are 1M the standard concentration

= $3.526 V - \frac{0.0591}{2} \log \left(\frac{1}{1} ^ 2\right)$
= 3.526V