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Given that E°red = -0.40 V for Cd2+/Cd at 25°C, find E° and E for the concentration cell expressed using shorthand notation below.

Cd(s) ∣ Cd2+(1.0 × 10-5 M) ∣∣ Cd2+(0.100 M) ∣ Cd(s)

E° = 0.00 V and E = +0.12 V


E° = 0.00 V and E = +0.24 V


E° = -0.40 V and E = -0.16 V


E° = -0.40 V and E = -0.28 V

Respuesta :

pstnonsonjoku

From the information in the question, the  E° and E for the cell is 0.00 V and 0.12 V.

Using the Nernst equation;

Ecell =  E°cell - 0.0592/n log Q

We know that  E°cell = 0.00 V since the anode and cathode are both made up of cadmium.

Substituting the given values into the Nernst equation;

Ecell = 0.00 V - 0.0592/2 log (1.0 × 10-5 M/0.100 M)

Ecell =  0.00 V - 0.0296 log(1 × 10^-4)

Ecell = 0.12 V

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