Using other reducing agents

Suppose you replaced zinc as the reducing agent by tin. How far would the set of reductions go this time?

Here are the E° values again:

. . . and here is the tin value:

For each reduction to happen, the vanadium reaction has to have the more positive E° value because we want it to go to the right. That means that the tin must have the more negative value.

• In the first vanadium equation (from +5 to +4), the tin value is more negative. That works OK.
• In the second vanadium equation (from +4 to +3), the tin value is again the more negative. That works as well.
• But in the final vanadium reaction (from +3 to +2), tin no longer has the more negative E° value. Tin won't reduce vanadium(III) to vanadium(II).