The figure shows the behavior of Pt and Ag cathodes at which reduction of H3O to H2(g) occurs. Explain why the two curves are not superimposed. Pt Ag 0.5㎂ -0.2-0.3-0.4-0.6-0.7-0.8-0.9 E(V vs. S.C.E....

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The figure shows the behavior of Pt and Ag cathodes at which reduction of H3O to H2(g) occurs. Explain why the two curves are not superimposed.

Pt Ag 0.5㎂ -0.2-0.3-0.4-0.6-0.7-0.8-0.9 E(V vs. S.C.E.)

science chemistry

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Answer 1

Answer #1

Reduction takes place at Ag electrode, because it has more negative reduction potential value. We know that oxidation takes place at +ve E value but reduction takes place at -ve E value.

So, here Pt electrode act as anode and Ag electrode act as cathode. HenceH₃O to H₂ reduction takes place at Ag electrode.

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Answer 2

The figure shows the behavior of Pt and Ag cathodes at which reduction of H3O to H2(g) occurs. Explain why the two curves are not superimposed. Pt Ag 0.5㎂ -0.2-0.3-0.4-0.6-0.7-0.8-0.9 E(V vs. S.C.E....

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The figure shows the behavior of Pt and Ag cathodes at which reduction of H3O to H2(g) occurs. Explain why the two curves are not superimposed. Pt Ag 0.5㎂ -0.2-0.3-0.4-0.6-0.7-0.8-0.9 E(V vs. S.C.E....

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Use the graph to estimate K. 1.0 0.75 Yo₂ 0.5 0.25 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 [L] (M)

By application of the impulse-invariant transformation of H (s), the following transfer functions have been obtained. Find H (s), assuming T = 0.1 s: H1(z) = 2z (z − e−0.2)(z − e−0.1) H2(z) = z − e−0.6 (z − e−0.5)(z − e−0.4) H3(z) = z (z − 0.9)(z − 0.3)

The figure shows the behavior of Pt and Ag cathodes at which reduction of H3O to H2(g) occurs. Explain why the two curves are not superimposed. Pt Ag 0.5㎂ -0.2-0.3-0.4-0.6-0.7-0.8-0.9 E(V vs. S.C.E....

Homework Answers

Add Answer to: The figure shows the behavior of Pt and Ag cathodes at which reduction of H3O to H2(g) occurs. Explain why the two curves are not superimposed. Pt Ag 0.5㎂ -0.2-0.3-0.4-0.6-0.7-0.8-0.9 E(V vs. S.C.E....

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Use the graph to estimate K. 1.0 0.75 Yo₂ 0.5 0.25 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 [L] (M)

By application of the impulse-invariant transformation of H (s), the following transfer functions have been obtained. Find H (s), assuming T = 0.1 s: H1(z) = 2z (z − e−0.2)(z − e−0.1) H2(z) = z − e−0.6 (z − e−0.5)(z − e−0.4) H3(z) = z (z − 0.9)(z − 0.3)

Add Answer to:
The figure shows the behavior of Pt and Ag cathodes at which reduction of H3O to H2(g) occurs. Explain why the two curves are not superimposed. Pt Ag 0.5㎂ -0.2-0.3-0.4-0.6-0.7-0.8-0.9 E(V vs. S.C.E....