A chemist prepares a solution of silver(I) nitrate (AgNO3) by measuring out 269. micromol of silver(I) nitrate into a 300. mL vo
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Answer:
See explanation and image attached
Explanation:
The standard cell potential at 298 K is given by;
E°cathode - E°anode
Hence;
E°cell = 0.34 V - (-0.76 V)
E°cell = 0.34 V + 0.76 V
E°cell = 1.1 V
To reduce Zn^2+ to Zn then Zn must be the cathode, hence;
E°cell = (-0.76 V) - 0.34 V
E°cell = -1.1 V
