© Macmillan Learning E = Consider the reactions. Zn2++ 4 NH3 Zn(NH₂)²+ = 5.01 × 108 E°=-0.762 V Zn2+ + 2eZn(s) Assuming there is negligible current and nearly all Zn² + is in the form Zn(NH3)²+, what cathode potential (vs. S.H.E.) would be required to reduce 99.99% of the Zn² + from a solution containing 0.12 M Zn² + in 1.0 M of ammonia? Assume T = 298 K. V

Fundamentals Of Analytical Chemistry
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Chapter18: Introduction To Electrochemistry
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© Macmillan Learning
E =
Consider the reactions.
Zn2++ 4 NH3 Zn(NH₂)²+
= 5.01 × 108
E°=-0.762 V
Zn2+ + 2eZn(s)
Assuming there is negligible current and nearly all Zn² + is in the form Zn(NH3)²+, what cathode potential (vs. S.H.E.) would be
required to reduce 99.99% of the Zn² + from a solution containing 0.12 M Zn² + in 1.0 M of ammonia? Assume T = 298 K.
V
Transcribed Image Text:© Macmillan Learning E = Consider the reactions. Zn2++ 4 NH3 Zn(NH₂)²+ = 5.01 × 108 E°=-0.762 V Zn2+ + 2eZn(s) Assuming there is negligible current and nearly all Zn² + is in the form Zn(NH3)²+, what cathode potential (vs. S.H.E.) would be required to reduce 99.99% of the Zn² + from a solution containing 0.12 M Zn² + in 1.0 M of ammonia? Assume T = 298 K. V
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