Exercise 14.87

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Rucha Kulkarni 2A
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Exercise 14.87

Postby Rucha Kulkarni 2A » Sun Feb 11, 2018 10:28 pm

Does anyone know how to solve question 87 on the homework?

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Re: Exercise 14.87

Postby CameronJohari1J » Sun Feb 11, 2018 11:54 pm

The E of the cathode is -.76V and the E of the anode is unknown. The E of the cell overall is .16V. Set E of cell to E cathode minus E anode. Solving for E of cathode results in -.92V for M(s) to M4+

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