The standard electrode potential corresponding to the reaction ` Au^(3+) (aq) + 3e^(-) rarr Au (s)` is `1. 42 V`. This implies that .
(i) gold dissolves in ` 1M HCl`
(ii) metallic gole will be precipitated on passing hydrogen gas through gold salt solution
(iii) gold does not dissolve in ` 1M HCl` solution
(iv) metallic gold will not be precipitated on passing hydrogen gas through gold salt solution.

Options

A. (i), (ii)
B. (i), (iv)
C. (ii), (iv)
D. (ii), (iii)

Answer

D

Explanation

Gold will dissolve in `1 M HCl` if it gets oxidized and ` H+` ions fo `HCl` are reduced :
` Au(s) + 3H^(+) (aq.) rarr Au^(3+) (aq.) + 3/2 H_2(g) `
But this is not possible because ` E_(Au//Au^(3+))^(Ө) (-1.42 V)` is less than ` E_(H^+//H_(2))^Ө (0.00V)`.
In other words, the standard reduction potential of gold is greater than zero . Therefore, the reduced from of gold, i.e., Au (s) is mor stable than the reduced from of hydrogen (i.e., `H_2`) i.e., gold cannot be oxidized by ` H^(+)` ions to give ` H_2`. Hence, gold does not dissolve in ` 1M HCl` solution.
Since ` E_(Au^(3+) //Au)^(Ө) (1.42 V)` is greater than ` E_(H^+ //H_2)^Ө` (0.00 V), hydrogen gas passed through the salt solution of gold will reduce auric ions and metallic gold will be percipitated.