The price of a binary option, ignoring interest rates, is basically the same as the CDF $\phi(S)$ (or $1-\phi(S)$ ) of the terminal probability distribution. Generally that terminal distribution will be lognormal from the Black-Scholes model, or close to it. Option price is
$$C = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$
for calls and
$$ P = e^{-rT} \int_0^K \psi(S_T) dS_T$$
for puts.
Volatility widens the distribution and, under the Black-Scholes model, shifts its mode a bit. Generally speaking, increased volatility will
Increase the density in the "payoff region" for out-of-the-money options, thereby increasing their theoretical value. Assuming your option was worth 0.30 due to probabilities and not high risk-free rates $ r $, more volatility will increase its value.
Increase the density in the "no-payoff region" for in-the-money options, thereby decreasing their theoretical value. An option now worth 0.70 will lose value, as the probability of ending outside the payoff region is increased.
As volatility $\sigma$ approaches $ \infty $, all option prices converge toward 0 for calls and 1 for puts. In Black-Scholes land, even though the term $ \frac{\log(S_0/K)}{\sigma \sqrt{T}} \to 0$ and the probability distribution is spreading out all the way to infinity on the positive as well as negative side of the exponential of its distribution, it concentrates lognormally on values less than any finite strike.
Therefore, out-of-the-money calls will take on a maximum value at some volatility that concentrates as much probability as possible below the strike before concentrating the distribution too close to zero.
Edit: A huge thank-you to @Veeken to pointing out that it is out-of-the-money calls, rather than puts, which take on a maximum theoretical value.
