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What will be the spontaneous reaction wh...

What will be the spontaneous reaction when the following half reactions are combined ?
(i) `Fe^(3+) + e^(-) to Fe^(2+)` , (ii) `MnO_(4)^(-) + 8 H^(+) + 5 e^(-) to Mn^(2+) + 4 H_(2) O , (E^(Theta) = + 1.49 V`)
What is the value of `E_(cell)^(Theta)` ?

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AI Generated Solution

To determine the spontaneous reaction when the given half-reactions are combined, we will follow these steps: ### Step 1: Identify the half-reactions and their standard reduction potentials - The first half-reaction is: \[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad (E^\circ = +0.77 \, \text{V}) \] - The second half-reaction is: ...
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In the given chemical reaction, colors of the Fe^(2+) and Fe^(3+) ions, are respectively : 5Fe^(2+) + MnO_4^(-) + 8 H^(+) to Mn^(2+) + 4H_2O + 5Fe^(3+)

In the given chemical reaction, colors of the Fe^(2+) and Fe^(3+) ions, are respectively : 5Fe^(2+) + MnO_4^(-) + 8 H^(+) to Mn^(2+) + 4H_2O + 5Fe^(3+)

The standard reduction for the following reactions are : Fe^(3+) + 3e^(-) rarr Fe with E^(@) = - 0.036 V Fe^(2+) + 2e^(-) rarr Fe with E^(@) = - 0.44 V What would be the standard electrode potential for the reaction Fe^(3+) + e^(-) rarr Fe^(2+) ?

Complete the following reactions. (i) MnO_4^(-)+2H_2O+3e^(-)to"_________"+4OH^(-) (ii) MnO_4^(-)+8H^(+)+5e^(-)to "__________" +4H_2O (iii) MnO_4^(-)+e^(-)to "_________"

MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H_(2)O , E^(@) = 1.51V MnO_(2) + 4H^(+) + 2e^(-) righarrow Mn^(2+) + 2H_(2)O E^(@) = 1.23V E_(MnO_(4)^(-)|MnO_(2)

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