Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in $$2.0$$ $$M{Fe}^{2}$$ and $$0.02M$$ $${Fe}^{3}$$ solution. Given $${E}_{{ Fe }^{ 3+ }/{ Fe }^{ 2+ }}^{o}=0.771V$$.

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Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in

2.0 M F e^{2}

and

0.02 M F E^{3}

solution. Given

E_{F e^{3} / F e^{+ 2}}^{0} = 0.771 V

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Calculate the cell potential for the following galvanic cell:
The first electrode consists of $${ Fe }^{ 3+ }|{ Fe }^{ 2+ }$$ couple in which $$\left[ { Fe }^{ 3+ } \right] =1M$$ and $$\left[ { Fe }^{ 2+ } \right] =\left[ 0.1M \right] $$
The second electrode consists of $$Mn{ O }_{ 4 }^{ - }|{ Mn }^{ 2+ }$$ couple in acidic solution in which
$$\left[ Mn{ O }_{ 4 } \right] =1\times { 10 }^{ -2 }M,\left[ { Mn }^{ 2+ } \right] =1\times { 10 }^{ -4 }M,\left[ { H }^{ + } \right] =1\times { 10 }^{ -3 }M\quad $$
$$\left( { E }_{ { Fe }^{ 3+ },{ Fe }^{ 2+ } }^{ o }=0.771V,{ E }_{ Mn{ O }_{ 4 }^{ - },{ Mn }^{ 2+ } }^{ o }=1.51V \right) $$

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Standard reduction potential of $$Fe^{3+} + e^{-} \rightarrow Fe^{2+}$$ is 0.77 V. What is the potential of a $$Pt$$ electrode immersed in an aqueous solution containing 2.0 M $$Fe^{2+}$$ and 0.2 M $$Fe^{3+}$$?

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The reduction of a half cell consisting of a Pt electrodes immersed in 1.5 M

{F e}^{2 +}

and 0.015 M

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(A) 0.652 V (B) 0.88 V (C) 0.710 V (D) 0.850 V

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