For Zn2- - Zn- E- - -0-76 V- for Ag-Ag- E- - 0-799 V- The correct statement is-

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For $Z n^{2 +} / Z n, E^{\circ} = - 0.76 V$ , for $A g^{+} / A g, E^{\circ} = 0.799 V$ . The correct statement is:

the reaction Zn getting reduced Ag getting oxidized is spontaneous

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Zn undergoes reduction and Ag is oxidized

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Zn undergoes oxidation Ag gets reduced

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{Z n}^{2 +} / Z n

E^{\circ} = - 0.76 V

{A g}^{+} / A g E^{\circ} = 0.799 V

{Z n}^{2 +} / Z n, E^{o} = - 0.76 V, f o r {A g}^{+} / A g, E^{o} = 0.799 V

△ G^{\circ}

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ Z n^{2 +} (a q) + 2 A g (s) + 2 O H^{-} (a q)

E_{A g^{+} / A g}^{\circ} = + 0.80 V

E_{Z n^{2 +} / Z N}^{\circ} = - 0.76 V

E^{\circ}

2 A g + Z n^{2 +} \to 2 A g^{+} + Z n

E_{A g^{+} / A g}^{\circ} = 0.80 V, E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V

A g (s) A g I O_{3} (s) | A g^{+} (x M), H I O_{3} (0.300 M) | ∣ ∣ Z n^{2 +} (0.175 M) ∣ ∣ Z n (s)

K_{S P} = 3.02 \times 10^{- 8}

A g I O_{3} (s)

K_{a} = 0.162

H I O_{3}, E^{\circ} (Z n^{2 +} / Z n) = - 0.76 V, E^{\circ} (A g / A g^{+} = - 0.8 V)

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