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

Question

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#

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Answer 1 · 2017-11-12T16:34:20

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+#.