Answer:
0.477 V
Step-by-step explanation:
When a substance is gaining electrons, it's reducing, and when the substance loses electrons, it's oxidizing. In a galvanic cell, one substance oxides giving electrons for the other, which reduces. Then, the substance with higher reduction potential must reduce and the other must oxide.
E°cell = E°red(red) - E°red(oxid)
Where, E°red(red) is the reduction potential of the substance that reduces, and E°red(oxid) is the reduction potential of the substance that oxides. For the value given, Cu⁺² reduces, so:
E°cell = +0.337 - (-0.140)
E°cell = 0.477 V