A simple example helps us identify what is essential.
$$Y = C + \gamma X_1 + \varepsilon$$
where $C$ and $\gamma$ are parameters, $X_1$ is the score on the first instrument (or independent variable), and $\varepsilon$ represents unbiased iid error. Let the score on the second instrument be related to the first one via
$$X_1 = \alpha X_2 + \beta.$$
For example, scores on the second instrument might range from 25 to 75 and scores on the first from 0 to 100, with $X_1 = 2 X_2 - 50$. The variance of $X_1$ is $\alpha^2$ times the variance of $X_2$. Nevertheless, we can rewrite
$$Y = C + \gamma(\alpha X_2 + \beta) = (C + \beta \gamma) + (\gamma \alpha) X_2 + \varepsilon = C' + \gamma' X_2 + \varepsilon.$$
The parameters change, and the variance of the independent variable changes, yet the predictive capability of the model remains unchanged.
In general the relationship between $X_1$ and $X_2$ may be nonlinear. Which is a better predictor of $Y$ will depend on which has a closer linear relationship to $Y$. Thus the issue is not one of scale (as reflected by the variance of the $X_i$) but has to be decided by the relationships between the instruments and what they are being used to predict. This idea is closely related to one explored in a recent question about selecting independent variables in regression.
There can be mitigating factors. For instance, if $X_1$ and $X_2$ are discrete variables and both are equally well related to $Y$, then the one with larger variance might (if it is sufficiently uniformly spread out) allow for finer distinctions among its values and thereby afford more precision. E.g., if both instruments are questionnaires on a 1-5 Likert scale, both are equally well correlated with $Y$, and the answers to $X_1$ are all 2 and 3 and the answers to $X_2$ are spread among 1 through 5, $X_2$ might be favored on this basis.