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23 Nov 2019

Consider a galvanic cell based on the following two half reactions at 277 K:

Cu2+ + 2 e- â Cu Eo = 0.34 V

Cr3+ + e- â Cr2+ Eo = -0.41 V

Which of the following changes will increase the potential of the cell?

Decreasing the concentration of Cr2+ ions in solution

Increasing the concentration of Cr2+ ions in solution

Increasing the mass of the copper electrode

Adding equal volumes of water to both half reactions

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Related questions

Consider a cell based on the following line notation at 298 K:

Cu | Cu²⁺ (0.289 M) || Ni²⁺ (1.25 M) | Ni

Given:

Cu²⁺ + 2 e^- â Cu E^o = 0.34 V
Ni²⁺ + 2 e^- â Ni E^o = -0.23 V

How many of the following responses are true?

1. Using a Pt electrode in place of the Cu electrode will not change the potential of the cell
2. Reduction takes place at the cathode
3. Adding equal amounts of water to both half reactions will not change the potential of the cell
4. Increasing the concentration of the Ni²⁺ will decrease the potential of the cell
5. Increasing the concentration of the Cu²⁺ will decrease the concentration of the reaction

Consider a cell based on the following line notation at 298 K:

Cu | Cu2+ (0.253 M) || Ni2+ (1.35 M) |Ni

Given: Cu2+ + 2 e- â Cu Eo = 0.34 V

Ni2+ + 2 e- â Ni Eo = -0.23 V

How many of the following responses are true?

1. Increasing the concentration of the Ni2+ will decrease the potential of the cell

2. Decreasing the concentration of the Cu2+ will increase the concentration of the reaction

3. Using a Pt electrode in place of the Cu electrode will not change the potential of the cell

4. Reduction takes place at the cathode

5. Adding equal amounts of water to both half reactions will decrease the potential of the cell

I said 3 and it was wrong

A concentration cell is built based on the following half reactions by using two pieces of copper as electrodes, two Cu2+ solutions, 0.145 M and 0.405 M, and all other materials needed for a galvanic cell. What will the potential of this cell be when the cathode concentration of Cu2+ has changed by 0.031 M at 279 K?

Cu2+ + 2 e- â Cu Eo = 0.341 V

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