Four alkali metals $A, B, C$ and $D$ are having respectively standard reduction potentials as $3.05, 1.66, 0.40$ and $0.80$ V. Which one will be the most reducing agent?

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Solution

The correct option is A A
Reducing nature is directly proportional to standard reduction potentials.
Reducing agent $\propto$ negative S & P value.
So, among the A, B, C and D. A is the most reducing agent.
Hence, option A is correct.

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Similar questions

Q. Four alkali metals

A, B, C and D

are having respective standard electrode potential as

- 3.05, - 1.66, - 0.40

and

0.80 V

.
Which one will be the powerful reducing agent?

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A, B, C

and

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are having standard electrode potentials 3.05, -1.66, -0.40 and 0.80V respectively. Which one will be the most reducing agent?

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and

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(standard reduction potential) of three metals A, B and C are respectively + 0.5 V; – 3.0 and – 1.2 V respectively. The reducing powers of these metals are:

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Four alkali metals A,B,C and D are having respectively standard reduction potentials as 3.05,1.66,0.40 and 0.80 V. Which one will be the most reducing agent?

The correct option is A AReducing nature is directly proportional to standard reduction potentials.Reducing agent ∝ negative S & P value.So, among the A, B, C and D. A is the most reducing agent.Hence, option A is correct.

Four alkali metals $A, B, C$ and $D$ are having respectively standard reduction potentials as $3.05, 1.66, 0.40$ and $0.80$ V. Which one will be the most reducing agent?

The correct option is A A
Reducing nature is directly proportional to standard reduction potentials.
Reducing agent $\propto$ negative S & P value.
So, among the A, B, C and D. A is the most reducing agent.
Hence, option A is correct.