The standard electrode potentials of two half-cells are given below: Sn^4+(aq) + 2e^- Sn^2+(aq) E^ = +0.15 V Br2(aq) + 2e^- 2Br^-(aq) E^ = +1.09 V Both half-cells use inert platinum electrodes, since neither redox couple includes a solid metal. (a) Construct the cell diagram (ce…

Answer: ((a)) Pt(s) Sn^2+(aq), Sn^4+(aq) ∥ Br2(aq), Br^-(aq) Pt(s) ((b)) Ecell^ = (+1.09) - (+0.15) = +0.94 V ((c)) Yes: bromine can oxidise Sn^2+(aq) to Sn^4+(aq) under standard conditions, because E^ (Br2/Br^-) = +1.09 V is more positive than E^ (Sn^4+/Sn^2+) = +0.15 V, so Br2 has the greater tendency to be reduced; the resulting Ecell^ = +0.94 V is positive, so the reaction is feasible. ((d)) Br2(aq) + Sn^2+(aq) → 2Br^-(aq) + Sn^4+(aq). Part (a): Cell diagram To decide which half-cell is the cathode (reduction, positive electrode) and which is the anode (oxidation, negative electrode), compare the two E^ values: the half-cell with the more positive E^ is more readily redu…

A2 Chemistry 9701: Standard Cell Potentials, Cell Notation and Feasibility

The standard electrode potentials of two half-cells are given below:

$\text{Sn}^{4+}(aq) + 2e^- \rightleftharpoons \text{Sn}^{2+}(aq) \qquad E^{\ominus} = +0.15\ \text{V}$

$\text{Br}_2(aq) + 2e^- \rightleftharpoons 2\text{Br}^-(aq) \qquad E^{\ominus} = +1.09\ \text{V}$

Both half-cells use inert platinum electrodes, since neither redox couple includes a solid metal.

(a) Construct the cell diagram (cell notation) for the electrochemical cell formed from these two half-cells under standard conditions. [2]

(b) Calculate the standard cell potential, $E_{cell}^{\ominus}$ , for this cell. [1]

(c) State, with a reason based on the $E^{\ominus}$ values given above, whether aqueous bromine is able to oxidise $\text{Sn}^{2+}(aq)$ to $\text{Sn}^{4+}(aq)$ under standard conditions. [2]

(d) Write the overall ionic equation for the feasible reaction identified in (c). [2]

Part (a): Cell diagram

To decide which half-cell is the cathode (reduction, positive electrode) and which is the anode (oxidation, negative electrode), compare the two $E^{\ominus}$ values: the half-cell with the more positive $E^{\ominus}$ is more readily reduced, so it is the cathode.

$E^{\ominus}(\text{Br}_2/\text{Br}^-) = +1.09\ \text{V}$ is more positive $\implies$ cathode (reduction: $\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$ ).
$E^{\ominus}(\text{Sn}^{4+}/\text{Sn}^{2+}) = +0.15\ \text{V}$ is less positive $\implies$ anode (oxidation: $\text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^-$ ).

By convention the anode is written on the left and the cathode on the right, with each half-cell’s species listed in the order matched to its own direction of reaction (reduced form nearest the electrode on the oxidation side; oxidised form nearest the electrode on the reduction side), and a double line (||, the salt bridge) separating the two half-cells:

$\text{Pt(s)} \mid \text{Sn}^{2+}(aq), \text{Sn}^{4+}(aq) \parallel \text{Br}_2(aq), \text{Br}^-(aq) \mid \text{Pt(s)}$

Part (b): Standard cell potential

$E_{cell}^{\ominus} = E^{\ominus}(\text{cathode, reduction}) - E^{\ominus}(\text{anode, oxidation})$

$E_{cell}^{\ominus} = (+1.09) - (+0.15) = +0.94\ \text{V}$

Part (c): Feasibility of bromine oxidising $\text{Sn}^{2+}$

Since $E^{\ominus}(\text{Br}_2/\text{Br}^-) = +1.09\ \text{V}$ is more positive than $E^{\ominus}(\text{Sn}^{4+}/\text{Sn}^{2+}) = +0.15\ \text{V}$ , the $\text{Br}_2/\text{Br}^-$ couple has the greater tendency to accept electrons (be reduced), while the $\text{Sn}^{4+}/\text{Sn}^{2+}$ couple has the greater tendency to release electrons (i.e. $\text{Sn}^{2+}$ is oxidised). Combining them as calculated in (b) gives $E_{cell}^{\ominus} = +0.94\ \text{V}$ , which is positive, so the reaction

$\text{Br}_2(aq) + \text{Sn}^{2+}(aq) \rightarrow 2\text{Br}^-(aq) + \text{Sn}^{4+}(aq)$

is feasible under standard conditions. Yes, bromine can oxidise $\text{Sn}^{2+}(aq)$ to $\text{Sn}^{4+}(aq)$ .

Part (d): Overall ionic equation

Combine the two half-equations, checking that electrons cancel exactly:

$\text{Br}_2(aq) + 2e^- \rightarrow 2\text{Br}^-(aq)$ $\text{Sn}^{2+}(aq) \rightarrow \text{Sn}^{4+}(aq) + 2e^-$

Both transfer $2$ electrons, so adding them directly and cancelling the electrons gives:

$\text{Br}_2(aq) + \text{Sn}^{2+}(aq) \rightarrow 2\text{Br}^-(aq) + \text{Sn}^{4+}(aq)$

Check (independent recomputation): atoms (Br: $2 = 2$ ✓, Sn: $1 = 1$ ✓. Charge) left: $0 + (+2) = +2$ ; right: $2(-1) + (+4) = -2 + 4 = +2$ . Equal ✓. The equation is balanced.

Final answers

(a) $\text{Pt(s)} \mid \text{Sn}^{2+}(aq), \text{Sn}^{4+}(aq) \parallel \text{Br}_2(aq), \text{Br}^-(aq) \mid \text{Pt(s)}$
(b) $E_{cell}^{\ominus} = \boxed{+0.94\ \text{V}}$
(c) Yes, feasible: $E_{cell}^{\ominus} = +0.94\ \text{V} > 0$ .
(d) $\text{Br}_2(aq) + \text{Sn}^{2+}(aq) \rightarrow 2\text{Br}^-(aq) + \text{Sn}^{4+}(aq)$