Calculate the standard reduction potential of  PuO22+ to Pu(s), given the following standard reduction potentials.  PuO22+ + 4H+ +2e- => Pu4+ + 2H2O   E° = 1.000 V Pu4+ + e- => Pu3+   E° = 1.006 V Pu3+ + 3e- => Pu(s)   E° = -1.978 V PuO22+ + e- => PuO2+ + 2H2O    E° = -0.966 V Which option is the correct answer? -0.488 v 0.664 V 0.168 V -3.066 V 0.028 V

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