Question

Halide ions can be deposited at a silver anode, the reaction being

Ag1s2 1 X2 h AgX1s2 1 e2

Suppose that a cell was formed by immersing a silver anode in an analyte solution that was 0.0250 M Cl2,

Br2, and I2 ions and connecting the half-cell to a saturated calomel cathode via a salt bridge.

(a) Which halide would form first and at what potential? Is the cell galvanic or electrolytic?

(b) Could I2 and Br2 be separated quantitatively? (Take 1.00 3 1025

M as the criterion for quantitative

removal of an ion.) If a separation is feasible, what range of cell potential could be used?

(c) Repeat part (b) for I2 and Cl2.

(d) Repeat part (b) for Br2 and Cl2.

Short answer

part a - AgI forms first at a potential of 0.0300 V (galvanic cell).

part b -

AgBr does not form until Ecell= 0.076 V so the separation is feasible if the Ecelldoes is

kept above 0.076 V.

part c - I– could be separated from Cl– by maintaining the cell potential above –0.073 V.

part d - But AgCl forms at a cell potential of –0.073 V, so separation is not feasible.

Step-by-step solution

part a - given information We calculate the potentials at which the silver salts first form  For I–, Ecell = 0.244 – (–0.151 – 0.0592 log 0.0250) = 0.300 V  For Br–, Ecell  = 0.244 – (0.073 – 0.0592 log 0.0250) = 0.076 V  For Cl–, Ecell  = 0.244 – (0.222 – 0.0592 log 0.0250) = –0.073 V

part a - AgI forms first at a potential of 0.0300 V (galvanic cell).

part b - given information We calculate the potentials at which the silver salts first form  For I–, Ecell = 0.244 – (–0.151 – 0.0592 log 0.0250) = 0.300 V  For Br–, Ecell  = 0.244 – (0.073 – 0.0592 log 0.0250) = 0.076 V  For Cl–, Ecell  = 0.244 – (0.222 – 0.0592 log 0.0250) = –0.073 V 

part b -

When [I–] ≤10–5M

Ecell= 0.244 – (–0.151 – 0.0592 log 10–5) = 0.099 V

AgBr does not form until Ecell= 0.076 V so the separation is feasible if the Ecelldoes is

kept above 0.076 V.

part c - given information We calculate the potentials at which the silver salts first form  For I–, Ecell = 0.244 – (–0.151 – 0.0592 log 0.0250) = 0.300 V  For Br–, Ecell  = 0.244 – (0.073 – 0.0592 log 0.0250) = 0.076 V  For Cl–, Ecell  = 0.244 – (0.222 – 0.0592 log 0.0250) = –0.073 V

part c -

I– could be separated from Cl– by maintaining the cell potential above –0.073 V

part d - given informationWe calculate the potentials at which the silver salts first form  For I–, Ecell = 0.244 – (–0.151 – 0.0592 log 0.0250) = 0.300 V  For Br–, Ecell  = 0.244 – (0.073 – 0.0592 log 0.0250) = 0.076 V  For Cl–, Ecell  = 0.244 – (0.222 – 0.0592 log 0.0250) = –0.073 V

part d -

For [Br–] ≤10–5M

Ecell= 0.244 – (0.073 – 0.0592 log 10–5) = –0.125 V

But AgCl forms at a cell potential of –0.073 V, so separation is not feasible.