Question

What is the oxidation state of vanadium in the products? Explain how you got your answer

Answer must be based off of the two equations:
`V^(3+)+e^(-)rightleftharpoonsV^(2+)` `E^\Theta=-0.26V`
`2NO_(3)^(-)+4H^(+)+2e^(-)rightleftharpoons2NO_2+2H_2O` `E^\Theta=+0.81V`

Answer 1

The voltage of the cell will spontaneously be `1.07V`.

Generally, the more positive the standard reduction potential, the more likely that species will be reduced.

Since vanadium's reaction has a smaller reduction potential, it will generally become oxidized in a voltaic cell harboring these two electrodes.

Hence,

`V^(2+) rightleftharpoons V^(3+)+e^(-)`

Its oxidation state will be `3+`.