Question

Calculate the EMF of the cell at 298 K:

Pt, Br₂ | Br⁻ (0.01 M) || H⁺ (0.03 M) | H₂ (1 atm), Pt

Given: \[\ce{E^{\circ}_{\frac{1}{2}Br/Br^-}}\] = +1.08 V.

Answer 1

Given: The cell notation is Pt, Br₂ | Br⁻ (0.01 M) || H⁺ (0.03 M) | H₂ (1 atm), Pt

The standard reduction potential for bromine is

\[\ce{E^{\circ}_{\frac{1}{2}Br/Br^-}}\] = +1.08 V

The anode (oxidation) reaction is \[\ce{H2 (1 atm) -> 2H+ (0.03 M) + 2e-}\]

The standard oxidation potential for hydrogen is

\[\ce{E^{\circ}_{H_2/H^+} = E^{\circ}_{H^+/H_2}}\] = 0 V

The cathode (reduction) reaction is \[\ce{Br2 + 2e- -> 2Br- (0.01 M)}\]

We know that

\[\ce{E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}}\]

= 0 − 1.08

= −1.08 V

Overall cell reaction is

\[\ce{H2 (1 atm) + Br2 -> 2H+ (0.03 M) + 2Br- (0.01 M)}\]

The number of electrons transferred, n = 2

Determine the reaction quotient

\[\ce{Q = \frac{[H+]^2[Br-]^2}{P_{H_2}}}\]

= `((0.03)^2(0.01)^2)/(1)`

= `((9 xx 10^-4) xx (1 xx 10^-4))/(1)`

= 9 × 10⁻⁸

By using the Nernst equation

`E_"cell" = E_"cell"^circ - 0.0591/n log Q`

= `-1.08 - 0.0591/2 log(9 xx 10^-8)`

= −1.08 − 0.02955 × (7.0457)

= −1.08 − 0.208

= −1.288 V

∴ The EMF of the cell at 298 K is −1.288 V.