# How do you determine the concentrations of reagents in a galvanic cell when the cell "dies"?

Apr 9, 2017

["Ag"^"+"] = 2.22 × 10^"-18"color(white)(l) "mol/L"; "Ni"^"2+" = "1.50 mol/L"

#### Explanation:

The reactions are:

Anode: color(white)(m)"1×"["Ni(s)" → "Ni"^"2+""(aq) (1 mol/L") + "2e"^"-"]
Cathode: "2×"["Ag"^"+""(aq) (1 mol/L") + "e"^"-" → "Ag(s)"]

Cell: $\text{Ni(s)" + "2Ag"^"+""(aq)" ("1 mol/L") → "Ni"^"2+" "(aq)" ("1 mol/L") + "2Ag(s)}$

${E}_{\textrm{c e l l}}^{\circ} = \text{0.25 V + 0.7994 V" = "1.049 V}$

As the cell runs down, $\left[\text{Ni"^"2+}\right]$ increases and $\left[\text{Ag"^"+}\right]$ decreases.

This continues until

E_text(cell) = E_text(cell)^° - (RT)/(nF)lnQ = 1.049 V - (RT)/(nF)lnQ = 0

$\frac{R T}{n F} \ln Q = \text{1.049 V}$

(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298 color(red)(cancel(color(black)("K"))))/(2 × "96 485" color(red)(cancel(color(black)("J·V"^"-1"))))lnQ = 1.049 color(red)(cancel(color(black)("V")))

$\text{0.012 84} \ln Q = 1.049$

$\ln Q = \frac{1.049}{\text{0.012 84}} = 81.70$

$Q = {e}^{\text{81.70" = 3.04 × 10^"35}}$

Thus, when the cell is dead,

Q =(["Ni"^"2+"])/(["Ag"^"+"]^2) = 3.04 × 10^35

Almost all the $\text{Ag"^"+}$ ions (1 mol/L) will have disappeared.

Since the molar ratio of $\text{Ag"^"+":"Ni"^"2+} = 2 : 1$, the loss of 1 mol/L $\text{Ag"^"+}$ will cause the formation of 0.5 mol/L $\text{Ni"^"2+}$.

$\left[\text{Ni"^"2+}\right]$ will have increased to its maximum value of 1.5 mol/L.

1.5/( ["Ag"^"+"]^2) = 3.04 × 10^35

["Ag"^"+"]^2 = 1.5/(3.04 × 10^35) = 4.93 × 10^"-36"

["Ag"^"+"] = 2.22 × 10^"-18"color(white)(l) "mol/L"