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3 years ago

Why is the earts special

Chemistry

2 answers:

densk [106]3 years ago

6 0

The heart is special because it pumps blood to your body which makes you stay alive.

eduard3 years ago

5 0

If its heart you're trying to say its special because, it pumps blood to make you stay alive. If you're trying to say earth then that's special because it has the oxygen that we humans need hope this helps and please mark me the brainliest

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Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

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nikdorinn [45]

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