Quick questions on Galvanic cells and standard cell potentials (QCE Chemistry Unit 3)

The table lists half-reactions written as reductions, with associated E0 values. Highly positive E0 means the species is a strong oxidising agent (readily reduced). Highly negative E0 means the species is a strong reducing agent (readily oxidised; the reverse direction is favoured).

Daniell cell (Zn/Cu). E0(Cu2+/Cu) = +0.34 V; E0(Zn2+/Zn) = -0.76 V.

The conventional shorthand for a galvanic cell, used in QCAA EA and IA1:

The reduction potential table is a kind of "redox priority list". The species higher in the table (more positive E0) wins as the oxidising agent; the species lower (more negative E0) loses electrons.

A galvanic cell delivers current until equilibrium is reached. The reaction proceeds in the spontaneous direction, but the consequent change in ion concentrations (Zn2+ rises, Cu2+ falls) shifts the actual cell potential toward zero. When the cell is fully discharged, the half-cells are at electrochemical equilibrium and the voltage is zero. This is the chemistry of a flat battery.

E0(Cu2+/Cu) = +0.34 V; E0(Zn2+/Zn) = -0.76 V.

E0(Ag+/Ag) = +0.80 V; E0(Cu2+/Cu) = +0.34 V.

E0(Ag+/Ag) = +0.80 V; E0(Sn2+/Sn) = -0.14 V.

Define electrode potential and explain the role of the standard hydrogen electrode. [3 marks]

Using standard electrode potentials (Ag$^+$/Ag $= +0.80$; Mg$^{2+}$/Mg $= -2.36$), calculate $E^{\circ}_{cell}$ for the spontaneous combination and identify anode and cathode. [4 marks]

A galvanic cell has Zn|Zn$^{2+}$ and Sn$^{2+}$|Sn half-cells. (a) Write half-equations and overall cell reaction. (b) Calculate $E^{\circ}_{cell}$ ($E^{\circ}(\text{Sn}^{2+}/\text{Sn}) = -0.14$).