Ridge regression regularize the linear regression by imposing a penalty on the size of coefficients. Thus the coefficients are shrunk toward zero and toward each other. But when this happens and if the independent variables does not have the same scale, the shrinking is not fair. Two independent variables with different scales will have different contributions to the penalized terms, because the penalized term is a sum of squares of all the coefficients. To avoid such kind of problems, very often, the independent variables are centered and scaled in order to have variance 1.
[Later edit to answer to comment]
Suppose now that you have an independent variable $height$. Now, human height might be measured in inches or meters or kilometers. If measured in kilometers, than in standard linear regression I think it will give a much bigger coefficient term, than if measured in millimeters.
The penalization term with lambda is the same as expressing the square loss function with respect to the sum of squared coefficients less than or equal a given constant. That means, bigger lambda gives much space to the squared sum of coefficients, and lower lambda a smaller space. Bigger or smaller space means bigger or smaller absolute values of the coefficients.
By not using standardization, then to fit the model might require big absolute values of the coefficients. Of course, we might have a big coefficient value naturally, due to the role of the variable in the model. What I state is that this value might have an artificially inflated value due to not scaling. So, scaling also decreases the need for a big values of coefficients. Thus, the optimal value of lambda would be usually smaller, which corresponds to a smaller sum of squared values of coefficients.
