Ni(s) + 2Ag+ (1.0M) → Ni2+ (1.0 M) + 2Ag(s) given that Eø for Ni+2/Ni= -0.25, Eø for Ag+/Ag=0.80. Need the emf of the cell. The Eø should be 0.55 as Eø for Ni+2/Ni= -0.25 but this is an anodic reaction so SOP= -0.25

Question

Ni(s) + 2Ag+ (1.0M) → Ni2+ (1.0 M) + 2Ag(s) given that Eø for Ni+2/Ni= -0.25, Eø for Ag+/Ag=0.80. Need the emf of the cell. The Eø should be 0.55 as Eø for Ni+2/Ni= -0.25 but this is an anodic reaction so SOP= -0.25

Aqeel ahmed · Answer

E° of cell = 0.80-(-0.25) = 0.80+0.25 =1.05 Emf of cell =E°-0.059÷n log p÷ R =1.05 - 0.059÷2 ( log 1- 2 log1) =1.05 Becoz ( log 1- 2 log 1 ) is zero so , product of 0.059 and o is zero....

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Ni(s) + 2Ag+ (1.0M) → Ni2+ (1.0 M) + 2Ag(s) given that Eø for Ni+2/Ni= -0.25, Eø for Ag+/Ag=0.80. Need the emf of the cell. The Eø should be 0.55 as Eø for Ni+2/Ni= -0.25 but this is an anodic reaction so SOP= -0.25

Ni(s) + 2Ag+ (1.0M) → Ni2+ (1.0 M) + 2Ag(s) given that Eø for Ni+2/Ni= -0.25, Eø for Ag+/Ag=0.80. Need the emf of the cell.
The Eø should be 0.55 as Eø for Ni+2/Ni= -0.25 but this is an anodic reaction so SOP= -0.25

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Ni(s) + 2Ag+ (1.0M) → Ni2+ (1.0 M) + 2Ag(s) given that Eø for Ni+2/Ni= -0.25, Eø for Ag+/Ag=0.80. Need the emf of the cell. The Eø should be 0.55 as Eø for Ni+2/Ni= -0.25 but this is an anodic reaction so SOP= -0.25

Ni(s) + 2Ag+ (1.0M) → Ni2+ (1.0 M) + 2Ag(s) given that Eø for Ni+2/Ni= -0.25, Eø for Ag+/Ag=0.80. Need the emf of the cell.The Eø should be 0.55 as Eø for Ni+2/Ni= -0.25 but this is an anodic reaction so SOP= -0.25

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Ni(s) + 2Ag+ (1.0M) → Ni2+ (1.0 M) + 2Ag(s) given that Eø for Ni+2/Ni= -0.25, Eø for Ag+/Ag=0.80. Need the emf of the cell.
The Eø should be 0.55 as Eø for Ni+2/Ni= -0.25 but this is an anodic reaction so SOP= -0.25