Question

vandard electrode potential of an electrode is greater than zero,
then we can infer that its
(A) reduced form is more stable compared to hydrogen gas.
(b) oxidised form is more stable compared to hydrogen gas.
(C) reduced and oxidised forms are equally stable.
(D) reduced form is less stable than the hydrogen gas.
mm​

Answer 1

Answer:

(A)

Explanation:

that

These two reaction can occur :-

${m}^{n + } - {m}^{1} or \: {h}^{ + } - {h2}^{}$

Since, the electrode potential for the first reaction is greater than zero. It will occur. Hence, it shows that the reduced form is more stable than H2.

Answer 2

We can infer that its, A) Reduced form is more stable compared to Hydrogen gas.

We know,

• As the Standard electrode potential increases more positively, the more it is a stronger oxidizing agent than Hydrogen.
• Which means that it is easier for that element to donate an electron than to receive it.
• It feels more stable in oxidizing another element, and itself getting reduced..
• This implies, it feels more stability in its reduced form.
• Therefore, if its SEP is more than Zero (SEP of Hydrogen),             then  its reduced form is more stable compared to Hydrogen gas.