# What is the voltage of a galvanic cell made with silver and nickel?

Mar 11, 2016

${E}_{c e l l}^{\circ} = + 1.05 \text{V}$

#### Explanation:

List the standard electrode potentials from -ve to +ve:

" "E^@("V")

$\stackrel{\textcolor{b l u e}{\leftarrow}}{\textcolor{w h i t e}{\times \times \times \times \times \times \times}}$
$N {i}_{\left(a q\right)}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s N {i}_{\left(s\right)} \text{ } - 0.25$

$A {g}_{\left(a q\right)}^{+} + e r i g h t \le f t h a r p \infty n s A {g}_{\left(s\right)} \text{ } + 0.8$

$\stackrel{\textcolor{red}{\rightarrow}}{\textcolor{w h i t e}{\times \times \times \times \times \times \times}}$
You can see that the $\text{Ag"^+"/""Ag}$ 1/2 cell has the most +ve ${E}^{\circ}$ value so this will take in electrons and move left to right.

The $\text{Ni"^(2+)"/""Ni}$ 1/2 cell will, therefore, take in these electrons and move right to left.

The voltage of the cell is an empirically measured quantity so must always have a +ve value. So to calculate ${E}_{c e l l}^{\circ}$ subtract the least +ve ${E}^{\circ}$ from the most +ve ${E}^{\circ}$ value$\Rightarrow$

${E}_{c e l l}^{\circ} = + 0.8 - \left(- 0.25\right) = + 1.05 \text{V}$