The E^ values for the half-cells Fe/Fe^ 2+ and Cu/Cu^ 2+ are -0.44 V and +0.32 V respectively. Then

JEE Main Chemistry question

The values for the half-cells and are V and V respectively. Then

(A) oxidises FeCorrect
(B)Cu oxidises
(C)Cu reduces
(D) oxidises

Answer: (A)

Solution

Since Fe has the lower (more negative) reduction potential, it is more easily oxidized than Cu; equivalently, (with the higher reduction potential) is a stronger oxidizing agent

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-143753