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The magnitude ( but not the sign ) of th...

The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:`
`Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V`
`X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V`
When the two half cells of `X` and `Y `are connected to construct a cell, electrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`.
If standard emf `(E^(c-))` of a half cell `Y^(2)|Y^(o+)` is `0.15V`, the standard emf of the half cell `Y^(o+)|Y` will be
(a)`0.19V`
(b)`0.53V`
(c)`0.49V`
(d)`0.64V`

A

`0.19V`

B

`0.53V`

C

`0.49V`

D

`0.64V`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the standard emf of the half-cell reaction \( Y^+ \leftrightarrow Y \) given the standard emf for the half-cell reaction \( Y^{2+} + 2e^- \leftrightarrow Y \) is \( 0.34 \, V \) and the standard emf for the half-cell reaction \( Y^{2+} + e^- \leftrightarrow Y^+ \) is \( 0.15 \, V \). ### Step-by-Step Solution: 1. **Identify the Given Data**: - Standard reduction potential for \( Y^{2+} + 2e^- \leftrightarrow Y \) is \( E_1 = 0.34 \, V \). - Standard reduction potential for \( Y^{2+} + e^- \leftrightarrow Y^+ \) is \( E_2 = 0.15 \, V \). 2. **Write the Half-Reactions**: - For the first half-reaction: \[ Y^{2+} + 2e^- \leftrightarrow Y \quad (E_1 = 0.34 \, V) \] - For the second half-reaction: \[ Y^{2+} + e^- \leftrightarrow Y^+ \quad (E_2 = 0.15 \, V) \] 3. **Determine the Required Half-Reaction**: - We need to find the standard reduction potential for the half-reaction: \[ Y^+ + e^- \leftrightarrow Y \] - Let's denote the standard reduction potential for this half-reaction as \( E_3 \). 4. **Relate the Potentials**: - The relationship between the potentials can be established using the following: \[ E_1 = E_2 + E_3 \] - Rearranging gives: \[ E_3 = E_1 - E_2 \] 5. **Substituting the Values**: - Substitute the known values into the equation: \[ E_3 = 0.34 \, V - 0.15 \, V \] - Calculate \( E_3 \): \[ E_3 = 0.34 \, V - 0.15 \, V = 0.19 \, V \] 6. **Conclusion**: - The standard emf of the half-cell reaction \( Y^+ \leftrightarrow Y \) is \( 0.19 \, V \). ### Final Answer: The standard emf of the half-cell \( Y^+ \leftrightarrow Y \) is \( 0.19 \, V \).

To solve the problem, we need to find the standard emf of the half-cell reaction \( Y^+ \leftrightarrow Y \) given the standard emf for the half-cell reaction \( Y^{2+} + 2e^- \leftrightarrow Y \) is \( 0.34 \, V \) and the standard emf for the half-cell reaction \( Y^{2+} + e^- \leftrightarrow Y^+ \) is \( 0.15 \, V \). ### Step-by-Step Solution: 1. **Identify the Given Data**: - Standard reduction potential for \( Y^{2+} + 2e^- \leftrightarrow Y \) is \( E_1 = 0.34 \, V \). - Standard reduction potential for \( Y^{2+} + e^- \leftrightarrow Y^+ \) is \( E_2 = 0.15 \, V \). ...
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The magnitude ( but not the sign ) of the standard reduction potentials of two metals X and Y are : Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V When the two half cells of X and Y are connected to construct a cell, eletrons flow from X to Y . When X is connected to a standard hydrogen electrode (SHE) ,electrons flow from X to SHE . Given the following half cell : YI+e^(-) rarr Y-I^(c-): " "E^(c-)=-0.27 V Solubility product of the iodide salt YI is

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