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2 years ago

Which metal will react spontaneously with Cu2+ (aq) at 25°C?

Chemistry

2 answers:

Gwar [14]2 years ago

5 0

Answer:

Explanation:

The standard reduction potentials are

E°/V

Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42

Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85

Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80

Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34

Mg2+(aq) + 2e- ⟶ Mg(s); -2.38

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.

Mg is the only metal with a standard reduction potential lower than that of Cu, so

Only Mg will react spontaneously with Cu²⁺.

Oksi-84 [34.3K]2 years ago

3 0

Answer:

B. Mg

Explanation:

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2 years ago

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2 years ago

Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

2 years ago