Question

A disposable galvanic cell $Zn|Zn^{2\oplus}||Sn^{2\oplus}|Sn$ is produced using 1.0 mL of 0.5 M $Zn(NO_3)_2$ and 1.0 mL of 0.50 M $Sn(NO_3)_2$. It is needed to power a pace-maker that draws a constant current of $10^{-6}$ Amp to run it and requires atleast 0.50 V to function. Calculate the value of $[Zn^{2\oplus}]$ when cell reaches 0.5 V at 298 K.

(Given : $E^\circ (Zn^{2\oplus}|Zn) = -0.76V; E^\circ (Sn^{2\oplus}|Sn) = -0.14 \text{ V}$).

Answer 1

0.9999 M

Answer 2

The given galvanic cell is $Zn|Zn^{2\oplus}||Sn^{2\oplus}|Sn$.

The standard reduction potentials are: $E^\circ (Zn^{2\oplus}|Zn) = -0.76 \text{ V}$ $E^\circ (Sn^{2\oplus}|Sn) = -0.14 \text{ V}$

1. Determine Anode, Cathode, and Overall Cell Reaction: Since $E^\circ (Zn^{2\oplus}|Zn)$ is more negative than $E^\circ (Sn^{2\oplus}|Sn)$, Zinc will be oxidized (anode) and Tin ions will be reduced (cathode).

Anode (Oxidation): $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$ Cathode (Reduction): $Sn^{2+}(aq) + 2e^- \rightarrow Sn(s)$

Overall Cell Reaction: $Zn(s) + Sn^{2+}(aq) \rightarrow Zn^{2+}(aq) + Sn(s)$ The number of electrons transferred, $n = 2$.

2. Calculate the Standard Cell Potential ($E^\circ_{cell}$): $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ $E^\circ_{cell} = E^\circ (Sn^{2\oplus}|Sn) - E^\circ (Zn^{2\oplus}|Zn)$ $E^\circ_{cell} = (-0.14 \text{ V}) - (-0.76 \text{ V})$ $E^\circ_{cell} = 0.62 \text{ V}$

3. Apply the Nernst Equation: The Nernst equation at 298 K is: $E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log_{10} Q$ Where $Q$ is the reaction quotient. For the given reaction: $Q = \frac{[Zn^{2+}]}{[Sn^{2+}]}$

We are given that the cell potential $E_{cell} = 0.50 \text{ V}$ at 298 K. Substitute the values into the Nernst equation: $0.50 \text{ V} = 0.62 \text{ V} - \frac{0.0592}{2} \log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]}$ $0.50 = 0.62 - 0.0296 \log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]}$

Rearrange to solve for the logarithm term: $0.0296 \log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]} = 0.62 - 0.50$ $0.0296 \log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]} = 0.12$ $\log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]} = \frac{0.12}{0.0296}$ $\log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]} \approx 4.054054$

Now, calculate the ratio $\frac{[Zn^{2+}]}{[Sn^{2+}]}$: $\frac{[Zn^{2+}]}{[Sn^{2+}]} = 10^{4.054054} \approx 11324.7$

4. Calculate the Concentrations: Initial concentrations: $[Zn^{2+}]_{initial} = 0.5 \text{ M}$ and $[Sn^{2+}]_{initial} = 0.5 \text{ M}$. Let $x$ be the change in concentration of $Sn^{2+}$ (moles/L) that has reacted when the cell potential is 0.5 V. According to the stoichiometry of the reaction, if $x$ moles/L of $Sn^{2+}$ are consumed, then $x$ moles/L of $Zn^{2+}$ are produced.

So, the concentrations at 0.5 V will be: $[Sn^{2+}] = 0.5 - x$ $[Zn^{2+}] = 0.5 + x$

Substitute these into the ratio: $\frac{0.5 + x}{0.5 - x} = 11324.7$

Solve for $x$: $0.5 + x = 11324.7 \times (0.5 - x)$ $0.5 + x = 5662.35 - 11324.7x$ $x + 11324.7x = 5662.35 - 0.5$ $11325.7x = 5661.85$ $x = \frac{5661.85}{11325.7} \approx 0.4999139 \text{ M}$

Now, calculate $[Zn^{2+}]$: $[Zn^{2+}] = 0.5 + x$ $[Zn^{2+}] = 0.5 + 0.4999139$ $[Zn^{2+}] = 0.9999139 \text{ M}$

Rounding to four significant figures, $[Zn^{2+}] \approx 0.9999 \text{ M}$.

Core Solution:

1. Identify anode (Zn) and cathode (Sn).
2. Write cell reaction: $Zn(s) + Sn^{2+}(aq) \rightarrow Zn^{2+}(aq) + Sn(s)$, $n=2$.
3. Calculate standard cell potential: $E^\circ_{cell} = E^\circ_{Sn^{2+}|Sn} - E^\circ_{Zn^{2+}|Zn} = -0.14 - (-0.76) = 0.62 \text{ V}$.
4. Apply Nernst equation at 298 K: $E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]}$.
5. Substitute $E_{cell} = 0.50 \text{ V}$: $0.50 = 0.62 - \frac{0.0592}{2} \log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]}$.
6. Solve for the concentration ratio: $\log_{10} \frac{[Zn^{2+}]}{[Sn^{2+}]} = \frac{0.12}{0.0296} \approx 4.054054$.
7. Calculate the ratio: $\frac{[Zn^{2+}]}{[Sn^{2+}]} = 10^{4.054054} \approx 11324.7$.
8. Let initial concentrations be $0.5 \text{ M}$. If $x$ moles/L of $Sn^{2+}$ react, then $[Sn^{2+}] = 0.5 - x$ and $[Zn^{2+}] = 0.5 + x$.
9. Set up equation: $\frac{0.5 + x}{0.5 - x} = 11324.7$.
10. Solve for $x$: $x \approx 0.4999139 \text{ M}$.
11. Calculate $[Zn^{2+}] = 0.5 + x = 0.5 + 0.4999139 \approx 0.9999139 \text{ M}$.
12. Rounding to four decimal places, $[Zn^{2+}] \approx 0.9999 \text{ M}$.