Question #9ba07

Question

Question #9ba07

1 Answer

Impact of this question

2937 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-18T02:33:25

$M n O_{2} (s) + 3 A g^{+} + 2 H_{2} O (l) \to M n O_{4}^{-} + 3 A g (s) ⏐ ↓ + 4 H^{+}$ $MnO_2(s) +3Ag^(+) + 2H_2O(l) rarr MnO_4^(-) + 3Ag(s)darr + 4H^(+)$

Are charge and mass balanced?

Explanation:

$Oxidation: M n (I V) \to M n (V I I)$ $"Oxidation: " Mn(IV) rarr Mn(VII)$

$M n O_{2} (s) + 2 H_{2} O (l) \to M n O_{4}^{-} + 4 H^{+} + 3 e^{-}$ $MnO_2(s) +2H_2O(l) rarr MnO_4^(-) + 4H^(+)+3e^-$ $(i)$ $(i)$ $; E^{\circ} = - 1.68 V$ $;E^@=-1.68V$

$Reduction: M n (I V) \to M n (V I I)$ $"Reduction: " Mn(IV) rarr Mn(VII)$

$A g^{+} + e^{-} \to A g (s)$ $Ag^(+) + e^(-) rarr Ag(s)$ $(i i)$ $(ii)$ $; E^{\circ} = 0.80 V$ $;E^@=0.80V$

$Overall: 1 \times (i) + 3 \times (i i)$ $"Overall: "1xx(i)+3xx(ii)$ ,

$M n O_{2} (s) + 3 A g^{+} + 2 H_{2} O (l) \to M n O_{4}^{-} + 3 A g (s) ⏐ ↓ + 4 H^{+}$ $MnO_2(s) +3Ag^(+) + 2H_2O(l) rarr MnO_4^(-) + 3Ag(s)darr + 4H^(+)$
$; E^{\circ} = - 0.78 V$ $;E^@=-0.78V$

I looked up the electrode potentials, and the reaction is NOT feasible. Reduction of $A g^{2 +}$ $Ag^(2+)$ would drive the reaction:

$A g^{2 +} + 2 e^{-} \to A g (s)$ $Ag^(2+) + 2e^(-) rarr Ag(s)$ $(i i i)$ $(iii)$ $; E^{\circ} = 1.99 V$ $;E^@=1.99V$

Science

Math

Humanities

... and beyond

Question #9ba07

1 Answer

Explanation:

Related questions

Impact of this question