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mash [69]
3 years ago
5

Given the reaction for the nickel-cadmium battery: 2NiOOH + Cd +2H2O -> 2Ni(OH)2 + Cd(OH)2 What species is oxidized during th

e discharge of the battery?
Chemistry
1 answer:
jok3333 [9.3K]3 years ago
8 0

Answer:

Cd is oxidized during the discharge of the battery

Explanation:

Based on the reaction:

2 NiOOH + Cd + 2H₂O → 2Ni(OH)₂ + Cd(OH)₂

And knowing Oxygen and hydrogen never change its charge, we must to find oxidation state of Ni and Cd before and after the reaction:

<em>Ni:</em>

In NiOOH: 2 O = -2*2 = -4 + 1H = +1, = -4 + 1 = -3. And as the molecule is neutral, Ni is 3+

In Ni(OH)₂: OH = -1. As there are 2 OH = -2. That means Ni is +2

The Ni is gaining one electron, that means is been reduced

<em>Cd:</em>

Cd before reaction is as pure solid with oxidation state = 0

Cd after the reaction is as Cd(OH)₂: 2 OH = -2. That means Cd is +2

The Cd is loosing 2 electrons, that means is the species that is oxidized.

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Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
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Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

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3 years ago
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Aluminium + Dioxygen = Ruby

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