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The standard reduction potentials at 295 K for the following half cells are given below: NO 3 - + 4 H + + 3 e - → NO g + 2 H 2 O E o = 0 . 97 V V 2 + aq + 2 e - → V s E o = - 1 . 19 V Fe 3 + aq + 3 e - → Fe s E o = - 0 . 04 V Ag + aq + e - → Ag s E o = 0 . 80 V Au 3 + aq + 3 e - → Au s E o = 1 . 40 V The number of metal(s) which will be oxidised by NO 3 - in aqueous solution is _____.

Correct answer

0

Step-by-step solution

For feasibility, check , E o cell = E cathode ( reduction ) o - E anode ( oxidation ) o > 0 The E o values when metal acts as anode and NO 3 - reaction is cathodic reaction. For Vanadium metal, E cell o = 0 . 97 + 1 . 19 = 2 . 16 V For Iron metal, E cell o = 0 . 97 + 0 . 04 = 1 . 01 V For silver metal, E cell o = 0 . 97 - 0 . 80 = 0 . 17 V For Gold metal, E cell o = 0 . 97 - 1 . 140 = - 0 . 17 V For electrodes having oxidation potential greater than - 0 . 97   V , E o cell > 0 ∴   Ag , &#160

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