The potential for the given cell Cr(s) | Cr3+(0.1M) | Fe2+(0.01M) | F…

The correct answer is Option (2) → 0.261V

The emf of a cell under non-standard conditions is given by the Nernst equation. At 298 K:

    \( \displaystyle E_{cell}=E^{0}_{cell}-\frac{0.059}{n}\log\frac{[\text{products}]}{[\text{reactants}]} \)

Here n is the number of electrons transferred in the balanced cell reaction, and only ions in solution appear in the log term (solids are taken as 1).

Chromium is the anode (left) and iron the cathode (right):

    \( \displaystyle \text{Anode: }\ce{Cr -> Cr^3+ + 3e^-}\quad(\times 2) \)

    \( \displaystyle \text{Cathode: }\ce{Fe^2+ + 2e^- -> Fe}\quad(\times 3) \)

    \( \displaystyle \text{Cell: }\ce{2Cr + 3Fe^2+ -> 2Cr^3+ + 3Fe},\quad n=6 \)

Substituting \( \ce{[Cr^{3+}]} \) = 0.1 M and \( \ce{[Fe^{2+}]} \) = 0.01 M:

    \( \displaystyle E_{cell}=0.30-\frac{0.059}{6}\log\frac{[\ce{Cr^3+}]^2}{[\ce{Fe^2+}]^3} \)

    \( \displaystyle E_{cell}=0.30-\frac{0.059}{6}\log\frac{(10^{-1})^2}{(10^{-2})^3}=0.30-\frac{0.059}{6}\log 10^{4} \)

    \( \displaystyle E_{cell}=0.30-0.00983\times4=0.30-0.039 \)

    \( \displaystyle E_{cell}=0.261\ \mathrm{V} \)

Shortcut: The log term is positive (the ratio is 10⁴), so E_cell must be a little less than 0.30 V. Only 0.261 V fits.

Note: It seems that there is a mistake in the original NTA paper. In the cell notation, the two half-cells are separated by a single vertical line (|); it should be a double vertical line (||), which stands for the salt bridge.